
If \[5{{x}^{2}}+\lambda {{y}^{2}}=20\] represents a rectangular hyperbola, then $\lambda$ equals
A.\[5\]
B. \[4\]
C. \[-5\]
D. None of these.
Answer
161.7k+ views
Hint: To solve this question we will use the general equation of hyperbola $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ and the condition of rectangular hyperbola which is $a=b$. We will first use the relation $a=b$ and substitute it in the general equation of hyperbola and derive an equation. We will then compare the derived equation with the general equation of hyperbola and derive the value of $\lambda$.
Complete step by step solution: We are given an equation \[5{{x}^{2}}+\lambda {{y}^{2}}=20\] which represents a rectangular hyperbola and we have to derive the value of $\lambda $. We know that the equation of hyperbola is $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ and the condition for a rectangular hyperbola is when $a=b$.
Now we will use the relation $a=b$ and substitute it in equation $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ and derive an equation.
$\begin{align}
& \dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1 \\
& \dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{a}^{2}}}=1 \\
& {{x}^{2}}-{{y}^{2}}={{a}^{2}}...........(i)
\end{align}$
First we will simplify the equation \[5{{x}^{2}}+\lambda {{y}^{2}}=20\]by dividing the equation by $5$.
\[\begin{align}
& {{x}^{2}}+\dfrac{\lambda }{5}{{y}^{2}}=4 \\
& {{x}^{2}}+\dfrac{\lambda }{5}{{y}^{2}}={{2}^{2}} \\
\end{align}\]
We will now compare the equation (i) with the equation \[{{x}^{2}}+\dfrac{\lambda }{5}{{y}^{2}}={{2}^{2}}\].
$\begin{align}
& \dfrac{\lambda }{5}=-1 \\
& \lambda =-5
\end{align}$
The value of $\lambda $ is $\lambda =-5$ when \[5{{x}^{2}}+\lambda {{y}^{2}}=20\] represents a rectangular hyperbola. Hence the correct option is (C).
Note: A rectangular hyperbola can be defined as a type of hyperbola in which the length of the transverse axis that is $2a$ and length of the conjugate axis that is $2b$ are of equal length. A hyperbola is said to be rectangular when \[a=b\] so its standard equation is ${{x}^{2}}-{{y}^{2}}={{a}^{2}}$.
The eccentricity of a rectangular hyperbola is $\sqrt{2}$ and the asymptotes of the rectangular hyperbola are perpendicular to each other. The coordinates of the parametric form of representation of a hyperbola is $x=a\sec \theta ,\,y=a\tan \theta $. The conjugate of a rectangular hyperbola is also a rectangular hyperbola.
Complete step by step solution: We are given an equation \[5{{x}^{2}}+\lambda {{y}^{2}}=20\] which represents a rectangular hyperbola and we have to derive the value of $\lambda $. We know that the equation of hyperbola is $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ and the condition for a rectangular hyperbola is when $a=b$.
Now we will use the relation $a=b$ and substitute it in equation $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ and derive an equation.
$\begin{align}
& \dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1 \\
& \dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{a}^{2}}}=1 \\
& {{x}^{2}}-{{y}^{2}}={{a}^{2}}...........(i)
\end{align}$
First we will simplify the equation \[5{{x}^{2}}+\lambda {{y}^{2}}=20\]by dividing the equation by $5$.
\[\begin{align}
& {{x}^{2}}+\dfrac{\lambda }{5}{{y}^{2}}=4 \\
& {{x}^{2}}+\dfrac{\lambda }{5}{{y}^{2}}={{2}^{2}} \\
\end{align}\]
We will now compare the equation (i) with the equation \[{{x}^{2}}+\dfrac{\lambda }{5}{{y}^{2}}={{2}^{2}}\].
$\begin{align}
& \dfrac{\lambda }{5}=-1 \\
& \lambda =-5
\end{align}$
The value of $\lambda $ is $\lambda =-5$ when \[5{{x}^{2}}+\lambda {{y}^{2}}=20\] represents a rectangular hyperbola. Hence the correct option is (C).
Note: A rectangular hyperbola can be defined as a type of hyperbola in which the length of the transverse axis that is $2a$ and length of the conjugate axis that is $2b$ are of equal length. A hyperbola is said to be rectangular when \[a=b\] so its standard equation is ${{x}^{2}}-{{y}^{2}}={{a}^{2}}$.
The eccentricity of a rectangular hyperbola is $\sqrt{2}$ and the asymptotes of the rectangular hyperbola are perpendicular to each other. The coordinates of the parametric form of representation of a hyperbola is $x=a\sec \theta ,\,y=a\tan \theta $. The conjugate of a rectangular hyperbola is also a rectangular hyperbola.
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